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I can't understand why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$

Edit: They are equal, thanks for the help

The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.

$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.

Use induction to prove it:

• Show it is true for $x=1$




• Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$




• Then show that it is true for $n=k+1$ by using the above formula.

Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}

You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),

Proof

Thus, the equality holds for all x in N*